Question Number 74225 by mathmax by abdo last updated on 20/Nov/19
![let p(x)=(1+jx)^n −(1−jx)^n with j=e^((i2π)/3) 1) determine the roots of p(x) and factorize P(x) inside C[x] 2) decompose the fraction F(x)=(1/(p(x)))](https://www.tinkutara.com/question/Q74225.png)
$${let}\:{p}\left({x}\right)=\left(\mathrm{1}+{jx}\right)^{{n}} −\left(\mathrm{1}−{jx}\right)^{{n}} \:\:{with}\:{j}={e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \\ $$$$\left.\mathrm{1}\right)\:{determine}\:{the}\:{roots}\:{of}\:{p}\left({x}\right)\:{and}\:{factorize}\:{P}\left({x}\right)\:{inside}\:{C}\left[{x}\right] \\ $$$$\left.\mathrm{2}\right)\:{decompose}\:{the}\:{fraction}\:{F}\left({x}\right)=\frac{\mathrm{1}}{{p}\left({x}\right)} \\ $$
Answered by mind is power last updated on 20/Nov/19
![p(x)=0⇒(1+jx)^n −(1−jx)^n =0 ⇔(((1+jx)/(1−jx)))^n =1 &x#(1/j) ((1+jx)/(1−jx))=e^((2ikπ)/n) ,x≠(1/j) k<n ⇔x=((e^((2ikπ)/n) −1)/(j(e^((2ikπ)/n) +1)))=(1/j) ((e^(i((kπ)/n)) −e^((−ikπ)/n) )/(e^(ik(π/n)) +e^(−((ikπ)/n)) ))=((i tan(((kπ)/n)))/j) x=(i/j).tan(((kπ)/(2n))) k∈[0,n−1] withe (k/n)≠2 if n=2s⇒k∈[1,2s−1]−{s} if n=2s+1⇒k∈[1,2s−1] p(x)=(1+jx)^n −(1−jx)^n p(x)=(jx)^n −(−jx)^n +n(jx)^(n−1) −n(−jx)^(n−1) ........ p(x)=(j^n +(−1)^(n+1) j^j )x^n +n(j^(n−1) +(−1)^n j^(n−1) )x^(n−1) if n=2s p is polynom degp=2s−1 if n=2s+1 degp=2s+1 n=2s p(x)= n(2j^(n−1) )Π_(k=0,k≠(n/2)) ^(n−1) (X−((itan(((kπ)/n)))/j)) if n=2s+1 p(x)=(2j^n )Π_(k=0) ^(n−1) (X−((itan(((kπ)/n)))/j)) 2)(1/(p(x))) we do it for n=2k+1 (1/(p(x)))=Σ_(k=0) ^(n−1) (a_k /((X−((itan(((kπ)/n)))/j)))) a_k =(1/(2j^n Π_(t=0,t#k) ^(n−1) (((i(tan(((tπ)/n))−tan(((kπ)/n)))/j)))) (1/(p(x)))=Σ_(k=0) ^(n−1) (1/(2j^n Π_(t=0,t#k) ^(n−1) (((i(tan(((tπ)/n))−tan(((kπ)/n)))/j)))).(1/((X−((itan(((kπ)/n)))/j)))) .](https://www.tinkutara.com/question/Q74257.png)
$${p}\left({x}\right)=\mathrm{0}\Rightarrow\left(\mathrm{1}+{jx}\right)^{{n}} −\left(\mathrm{1}−{jx}\right)^{{n}} =\mathrm{0} \\ $$$$\Leftrightarrow\left(\frac{\mathrm{1}+{jx}}{\mathrm{1}−{jx}}\right)^{{n}} =\mathrm{1}\:\&{x}#\frac{\mathrm{1}}{{j}} \\ $$$$\frac{\mathrm{1}+{jx}}{\mathrm{1}−{jx}}={e}^{\frac{\mathrm{2}{ik}\pi}{{n}}} ,{x}\neq\frac{\mathrm{1}}{{j}}\:\:\:\:{k}<{n} \\ $$$$\Leftrightarrow{x}=\frac{{e}^{\frac{\mathrm{2}{ik}\pi}{{n}}} −\mathrm{1}}{{j}\left({e}^{\frac{\mathrm{2}{ik}\pi}{{n}}} +\mathrm{1}\right)}=\frac{\mathrm{1}}{{j}}\:\:\:\frac{{e}^{{i}\frac{{k}\pi}{{n}}} −{e}^{\frac{−{ik}\pi}{{n}}} }{{e}^{{ik}\frac{\pi}{{n}}} +{e}^{−\frac{{ik}\pi}{{n}}} }=\frac{{i}\:{tan}\left(\frac{{k}\pi}{{n}}\right)}{{j}} \\ $$$${x}=\frac{{i}}{{j}}.{tan}\left(\frac{{k}\pi}{\mathrm{2}{n}}\right)\:\:\:{k}\in\left[\mathrm{0},{n}−\mathrm{1}\right]\:{withe}\:\:\frac{{k}}{{n}}\neq\mathrm{2} \\ $$$${if}\:{n}=\mathrm{2}{s}\Rightarrow{k}\in\left[\mathrm{1},\mathrm{2}{s}−\mathrm{1}\right]−\left\{{s}\right\} \\ $$$${if}\:{n}=\mathrm{2}{s}+\mathrm{1}\Rightarrow{k}\in\left[\mathrm{1},\mathrm{2}{s}−\mathrm{1}\right] \\ $$$${p}\left({x}\right)=\left(\mathrm{1}+{jx}\right)^{{n}} −\left(\mathrm{1}−{jx}\right)^{{n}} \\ $$$${p}\left({x}\right)=\left({jx}\right)^{{n}} −\left(−{jx}\right)^{{n}} +{n}\left({jx}\right)^{{n}−\mathrm{1}} −{n}\left(−{jx}\right)^{{n}−\mathrm{1}} …….. \\ $$$${p}\left({x}\right)=\left({j}^{{n}} +\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {j}^{{j}} \right){x}^{{n}} +{n}\left({j}^{{n}−\mathrm{1}} +\left(−\mathrm{1}\right)^{{n}} {j}^{{n}−\mathrm{1}} \right){x}^{{n}−\mathrm{1}} \\ $$$${if}\:{n}=\mathrm{2}{s}\:{p}\:{is}\:{polynom}\:{degp}=\mathrm{2}{s}−\mathrm{1} \\ $$$${if}\:{n}=\mathrm{2}{s}+\mathrm{1}\:\:{degp}=\mathrm{2}{s}+\mathrm{1} \\ $$$${n}=\mathrm{2}{s} \\ $$$${p}\left({x}\right)=\:\:{n}\left(\mathrm{2}{j}^{{n}−\mathrm{1}} \right)\underset{{k}=\mathrm{0},{k}\neq\frac{{n}}{\mathrm{2}}} {\overset{{n}−\mathrm{1}} {\prod}}\left({X}−\frac{{itan}\left(\frac{{k}\pi}{{n}}\right)}{{j}}\right) \\ $$$${if}\:{n}=\mathrm{2}{s}+\mathrm{1} \\ $$$${p}\left({x}\right)=\left(\mathrm{2}{j}^{{n}} \right)\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({X}−\frac{{itan}\left(\frac{{k}\pi}{{n}}\right)}{{j}}\right) \\ $$$$\left.\mathrm{2}\right)\frac{\mathrm{1}}{{p}\left({x}\right)} \\ $$$${we}\:{do}\:{it}\:{for}\:{n}=\mathrm{2}{k}+\mathrm{1} \\ $$$$\frac{\mathrm{1}}{{p}\left({x}\right)}=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{{a}_{{k}} }{\left({X}−\frac{{itan}\left(\frac{{k}\pi}{{n}}\right)}{{j}}\right)} \\ $$$${a}_{{k}} =\frac{\mathrm{1}}{\mathrm{2}{j}^{{n}} \underset{{t}=\mathrm{0},{t}#{k}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\frac{{i}\left({tan}\left(\frac{{t}\pi}{{n}}\right)−{tan}\left(\frac{{k}\pi}{{n}}\right)\right.}{{j}}\right)} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{{p}\left({x}\right)}=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{j}^{{n}} \underset{{t}=\mathrm{0},{t}#{k}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\frac{{i}\left({tan}\left(\frac{{t}\pi}{{n}}\right)−{tan}\left(\frac{{k}\pi}{{n}}\right)\right.}{{j}}\right)}.\frac{\mathrm{1}}{\left({X}−\frac{{itan}\left(\frac{{k}\pi}{{n}}\right)}{{j}}\right)} \\ $$$$ \\ $$$$. \\ $$
Commented by mathmax by abdo last updated on 21/Nov/19
$${thankx}\:{sir}. \\ $$